Prescaler for clock division in VHDL - Step-by-Step Execution

Concept Flow - Prescaler for clock division

Input Clock

↓

Counter increments

↓

Check if counter reached limit

↓

Wait for next clk

↓

Reset counter

↩Back to Counter increments

The prescaler counts input clock pulses until a set limit, then toggles the output clock and resets the count, effectively dividing the clock frequency.

Execution Sample

VHDL

process(clk)
  begin
    if rising_edge(clk) then
      if counter = N-1 then
        counter <= 0;
        clk_out <= not clk_out;
      else
        counter <= counter + 1;
      end if;
    end if;
  end process;

This code divides the input clock by toggling the output clock every N input clock cycles.

Execution Table

Step	clk Rising Edge	counter Before	Condition (counter = N-1)	Action Taken	counter After	clk_out After
1	Yes	0	No	counter <= 1	1	0
2	Yes	1	No	counter <= 2	2	0
3	Yes	2	No	counter <= 3	3	0
4	Yes	3	No	counter <= 4	4	0
5	Yes	4	No	counter <= 5	5	0
6	Yes	5	No	counter <= 6	6	0
7	Yes	6	No	counter <= 7	7	0
8	Yes	7	No	counter <= 8	8	0
9	Yes	8	No	counter <= 9	9	0
10	Yes	9	Yes	counter <= 0; clk_out <= not clk_out	0	1
11	Yes	0	No	counter <= 1	1	1
12	Yes	1	No	counter <= 2	2	1
13	Yes	2	No	counter <= 3	3	1
14	Yes	3	No	counter <= 4	4	1
15	Yes	4	No	counter <= 5	5	1
16	Yes	5	No	counter <= 6	6	1
17	Yes	6	No	counter <= 7	7	1
18	Yes	7	No	counter <= 8	8	1
19	Yes	8	No	counter <= 9	9	1
20	Yes	9	Yes	counter <= 0; clk_out <= not clk_out	0	0

💡 Process continues indefinitely on each rising edge of clk, toggling clk_out every N cycles.

Variable Tracker

Variable	Start	After 1	After 2	After 3	After 4	After 5	After 6	After 7	After 8	After 9	After 10	After 20
counter	0	1	2	3	4	5	6	7	8	9	0	0
clk_out	0	0	0	0	0	0	0	0	0	0	1	0

Key Moments - 3 Insights

Why does the output clock toggle only when the counter reaches N-1?

Because the prescaler counts from 0 to N-1, and toggling at N-1 ensures the output clock frequency is divided by N. See execution_table rows 10 and 20 where clk_out toggles.

What happens to the counter after it reaches N-1?

The counter resets to 0 immediately after reaching N-1, as shown in execution_table rows 10 and 20, to start counting the next cycle.

Why does the counter increment only on the rising edge of clk?

Because the process is triggered by rising_edge(clk), ensuring synchronization with the input clock. This is shown in the execution_table where each step corresponds to a rising edge.

Visual Quiz - 3 Questions

Look at the execution_table at step 10. What is the value of clk_out after this step?

A 1

B 0

C Undefined

D Same as before

💡 Hint

Check the 'clk_out After' column at step 10 in the execution_table.

At which step does the counter reset to 0 for the first time?

A Step 11

B Step 9

C Step 10

D Step 1

💡 Hint

Look at the 'counter After' column in the execution_table to find when it goes back to 0.

If N was changed from 10 to 5, how would the execution_table change?

A counter would count to 10 instead of 5

B clk_out toggles every 5 steps instead of 10

C clk_out would never toggle

D counter would increment by 2 each step

💡 Hint

Think about when the condition counter = N-1 becomes true and triggers clk_out toggle.

Concept Snapshot

Prescaler divides clock frequency by counting input pulses.
When counter reaches N-1, output clock toggles and counter resets.
Implemented with a counter incrementing on rising clock edges.
Output frequency = input frequency / (2 * N).
Useful for generating slower clocks from fast input clocks.

Full Transcript

This visual execution shows how a prescaler divides a clock signal in VHDL. The process increments a counter on each rising edge of the input clock. When the counter reaches N-1, it resets to zero and toggles the output clock signal. This toggling divides the input clock frequency by 2*N. The execution table traces each step, showing counter and output clock values. Key moments clarify why toggling happens at N-1 and why the counter resets. The quiz tests understanding of these steps and effects of changing N.

Practice

1. What is the main purpose of a prescaler in VHDL clock division?

easy

A. To generate random clock pulses

B. To multiply the input clock frequency

C. To reset the clock signal

D. To divide a fast input clock into a slower output clock

Solution

1. Step 1: Understand the function of a prescaler

   A prescaler reduces the frequency of a fast clock by dividing it to create a slower clock signal.

2. Step 2: Compare options with the prescaler function

   Only To divide a fast input clock into a slower output clock correctly describes this division function; others describe unrelated actions.

3. Final Answer:

   To divide a fast input clock into a slower output clock -> Option D

4. Quick Check:

   Prescaler divides clock = To divide clock [OK]

Hint: Prescaler always divides clock frequency, never multiplies [OK]

Common Mistakes:

• Thinking prescaler multiplies clock frequency
• Confusing prescaler with reset function
• Assuming prescaler generates random pulses

2. Which VHDL code snippet correctly declares a counter signal for a prescaler?

easy

A. signal counter : integer range 0 to 99 := 0;

B. signal counter : std_logic_vector(7 downto 0) := '0';

C. signal counter : boolean := false;

D. signal counter : integer := '0';

Solution

1. Step 1: Identify proper signal type for counting

   Counting requires an integer with a defined range and initial value, as in signal counter : integer range 0 to 99 := 0;.

2. Step 2: Check other options for correctness

   signal counter : std_logic_vector(7 downto 0) := '0'; uses std_logic_vector but initializes incorrectly with a single quote; signal counter : boolean := false; uses boolean which can't count; signal counter : integer := '0'; assigns a character to integer, which is invalid.

3. Final Answer:

   signal counter : integer range 0 to 99 := 0; -> Option A

4. Quick Check:

   Integer counter with range and init = Correct declaration [OK]

Hint: Use integer with range and initial value for counters [OK]

Common Mistakes:

• Using std_logic_vector without proper initialization
• Using boolean type for counting
• Assigning character literals to integer signals

3. Given this VHDL process for a prescaler with a max count of 3:

process(clk, reset)
begin
  if reset = '1' then
    counter <= 0;
    clk_out <= '0';
  elsif rising_edge(clk) then
    if counter = 3 then
      counter <= 0;
      clk_out <= not clk_out;
    else
      counter <= counter + 1;
    end if;
  end if;
end process;

What is the frequency division factor of clk_out relative to clk?

medium

A. 4

B. 8

C. 6

D. 3

Solution

1. Step 1: Understand counter behavior

   The counter counts from 0 to 3, so it counts 4 clock cycles before toggling clk_out.

2. Step 2: Calculate total division factor

   Since clk_out toggles every 4 input clocks, one full cycle of clk_out is 8 input clocks (4 up, 4 down).

3. Final Answer:

   8 -> Option B

4. Quick Check:

   Toggle every 4 clocks means full period 8 clocks [OK]

Hint: Multiply count by 2 for full output clock period [OK]

Common Mistakes:

• Using counter max value as division factor directly
• Forgetting clk_out toggles, doubling period
• Counting only half the clock cycles

4. Identify the error in this prescaler process code:

process(clk, reset)
begin
  if reset = '1' then
    counter <= 0;
    clk_out <= '0';
  elsif rising_edge(clk) then
    if counter = 4 then
      counter <= 0;
      clk_out <= not clk_out;
    else
      counter <= counter + 1;
    end if;
  end if;
end process;

Assuming counter is declared as integer range 0 to 3, what will happen?

medium

A. Counter never reaches 4, so clk_out never toggles

B. Counter resets correctly at 4

C. Runtime error due to counter overflow

D. clk_out toggles every clock cycle

Solution

1. Step 1: Check counter range and comparison

   The counter is declared with range 0 to 3, but the code checks if counter = 4, which is outside the range.

2. Step 2: Understand effect of out-of-range value

   Since counter can only count up to 3, it will never equal 4, so the condition is never true and clk_out never toggles.

3. Final Answer:

   Counter never reaches 4, so clk_out never toggles -> Option A

4. Quick Check:

   Counter max 3 but checks 4 = Condition never met [OK]

Hint: Check counter max matches comparison value [OK]

Common Mistakes:

• Ignoring counter range limits
• Assuming counter can reach out-of-range values
• Thinking clk_out toggles without counter reset

5. You want to design a prescaler in VHDL that divides a 50 MHz clock down to 1 kHz. Which counter range is closest to the correct choice?
Use the formula: division factor = 50,000,000 / 1,000 = 50,000.
Remember the output clock toggles every time the counter reaches max and resets.

hard

A. signal counter : integer range 0 to 99999 := 0;

B. signal counter : integer range 0 to 49999 := 0;

C. signal counter : integer range 0 to 24999 := 0;

D. signal counter : integer range 0 to 12499 := 0;

Solution

1. Step 1: Calculate division factor for toggling

   The full division factor is 50,000, but clk_out toggles every time counter reaches max, so counter max = 50,000 / 2 = 25,000.

2. Step 2: Choose closest counter range

   Since counting starts at 0, range should be 0 to 24,999 to count 25,000 cycles.

3. Final Answer:

   signal counter : integer range 0 to 24999 := 0; -> Option C

4. Quick Check:

   Divide by 2 for toggle count = 24999 max [OK]

Hint: Divide total count by 2 for toggle max count [OK]

Common Mistakes:

• Using full division factor as counter max
• Not dividing by 2 for toggle behavior
• Choosing incorrect range limits